1.Introduction

A.S. Janfada and  A. Abbaspour[1] showed that equation x^6+6z^3=y^6+6w^3 has infinite non-trivial primitive integer solutions.

We show that x^6 + az^2 = y^6 + aw^2 has a parametric solution.
     
     
2.Theorem
     
Diophantine equation x^6 + az^2 = y^6 + aw^2 has a parametric solution.

(x,y,z,w)=(2pta, 2qta, 4(t^2q^6-t^2p^6+a)t^2a^2, 4t^2a^2(-t^2q^6+t^2p^6+a))
a,p,q,t are arbitrary.
 
Proof.

x^6 + az^2 = y^6 + aw^2....................................................(1)

Substitute x=p, y=q to equation (1), we obtain

a(z+w)(z-w) = q^6-p^6......................................................(2)

Hence we obatin (z,w) =(  1/2(t^2q^6-t^2p^6+a)/(ta), -1/2(-t^2q^6+t^2p^6+a)/(ta)).

Finally, we obtain a parametric solution below.

(x,y,z,w)=(2pta, 2qta, 4(t^2q^6-t^2p^6+a)t^2a^2, 4t^2a^2(-t^2q^6+t^2p^6+a))

Q.E.D.



3.Reference

[1].A.S. Janfada and  A. Abbaspour, On Diophantine equations X^6+ 6Z^3 = Y^6± 6W^3,
    International Journal of Pure and Applied Mathematics, VOL:105, NO:4,2015